Plotting the exponential of 1/(y-x)

What is the best way to handle this situation?

f[x_, y_] = Exp[1/(x – y)];
Plot3D[f[x, y], {x, -5, 5}, {y, -5, 5},
MeshFunctions -> {#3 &},
AxesLabel -> Automatic]

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3

 

You can keep the original definition of the function and just add Exclusions -> x == y to the plot.
– Rahul
Oct 5 ’15 at 3:11

  

 

@m_goldberg: Seems to clean up the graph just fine: i.stack.imgur.com/hug0F.png
– Rahul
Oct 5 ’15 at 3:32

  

 

@Rahul Excellent! Thanks for the great help. 🙂
– David
Oct 5 ’15 at 3:49

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1 Answer
1

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I am presuming you want to eliminate the singularity. Perhaps this will work for you.

Clear[f]
f[x_, y_] /; Abs[x – y] > .05 := Exp[1/(x – y)]
Plot3D[f[x, y], {x, -5, 5}, {y, -5, 5},
ClippingStyle -> None,
MeshFunctions -> {#3 &},
AxesLabel -> Automatic]

Instead of .05, which I chose arbitrarily, you may find a smaller value more to your liking.

You could also do what Rahul suggested in a comment, but you will need to use Quiet to suppress the error messages.

Quiet @
Plot3D[Exp[1/(x – y)], {x, -5, 5}, {y, -5, 5},
Exclusions -> x == y,
ClippingStyle -> None,
MeshFunctions -> {#3 &},
AxesLabel -> Automatic]

This gives the a plot that is indistinguishable form the one shown above.

  

 

My helper is back! Thanks. However, I do not understand the function notation. Can you help me with what f[x_, y_] /; Abs[x – y] > .05 := Exp[1/(x – y)] does?
– David
Oct 5 ’15 at 3:07

1

 

@David. /; is the operator form of the functionCondition, which you should study. You will find very useful. In this case it makes f undefined in the region of the singularity, so Plot3D doesn’t try to plot there.
– m_goldberg
Oct 5 ’15 at 3:21

1

 

@David. I can make nothing of your “Ouch” comment. There is nothing in my answer that can destroy anything you have in your notebook. Clear[f] could delete previous definitions of f from your working kernel, which is what it should do, since they might interfere with the new definition. But it can’t destroy anything in the notebook itself.
– m_goldberg
Oct 5 ’15 at 3:25

  

 

My bad. I just have two functions f in the global workspace now.
– David
Oct 5 ’15 at 3:43

  

 

Two absolutely wonderful explanations. Thanks to both of you.
– David
Oct 5 ’15 at 3:49